删除所有正则表达式的情况下，开始用char'['和用char结束']'从字符串 [英] Deleting all regex instances starting with char '[' and ending with char ']' from a String

问题描述

我需要一个字符串键，删除所有正则表达式在它开始与字符'['和结束与字符']'。

I need to take a String and deleting all the regexes in it starting with character '[' and ending with character ']'.

现在我不知道如何解决这个问题。我试图字符串字符数组转换，然后从任起点'['，直到他结束']'，然后将其转换回<$ C $把空字符C>字符串使用的toString（）方法。

Now i don't know how to tackle this problem. I tried to convert the String to character array and then putting empty characters from any starting '[' till his closing ']' and then convert it back to a String using toString() method.

我的code：

char[] lyricsArray = lyricsParagraphElements.get(1).text().toCharArray();
                for (int i = 0;i < lyricsArray.length;i++)
                {
                    if (lyricsArray[i] == '[')
                    {
                        lyricsArray[i] = ' ';
                        for (int j = i + 1;j < lyricsArray.length;j++)
                        {
                            if (lyricsArray[j] == ']')
                            {
                                lyricsArray[j] = ' ';
                                i = j + 1;
                                break;
                            }
                            lyricsArray[j] = ' ';
                        }   
                    }
                }
                String songLyrics = lyricsArray.toString();
                System.out.println(songLyrics);

但在songLyrics的打印行，我得到怪异的东西，像

But in the print line of songLyrics i get weird stuff like

[C@71bc1ae4
[C@6ed3ef1
[C@2437c6dc
[C@1f89ab83
[C@e73f9ac
[C@61064425
[C@7b1d7fff
[C@299a06ac
[C@383534aa
[C@6bc168e5

我想有它的简单方法。任何帮助将是非常美联社preciated。

I guess there is a simple method for it. Any help will be very appreciated.

有关澄清：
转换ABCD [dsadsadsa] EFG [ADF％@ 1] D进入abcdefgd

推荐答案

或者干脆使用普通的前pression到的替换 \\\\所有 OCCURENCES [* \\\\] 一无所有：

Or simply use a regular expression to replace all occurences of \\[.*\\] with nothing:

String songLyrics = text.replaceAll("\\[.*?\\]", "");

其中，文本是ofcourse：

String text = lyricsParagraphElements.get(1).text();

---

这是什么 \\\\ [* \\\\] 意思？

---

What does \\[.*\\] mean?

的replaceAll 的第一个参数是描述一个普通的前pression的字符串。定期EX pression定义一个模式在字符串相匹配。

The first parameter of replaceAll is a string describing a regular expression. A regular expression defines a pattern to match in a string.

因此​​，让我们把它分解了：

So let's split it up:

\\\\ [完全匹配字符 [。由于 [有一个普通的前pression中的特殊意义，它需要进行转义（两次！）。

\\[ matches exactly the character [. Since [ has a special meaning within a regular expression, it needs to be escaped (twice!).

。匹配任何字符，用（<一本结合起来href=\"http://stackoverflow.com/questions/2301285/what-do-lazy-and-greedy-mean-in-the-context-of-regular-ex$p$pssions\">lazy)零或更多的运营商 * ，并且直到最后发现它会匹配任何字符：

. matches any character, combine this with the (lazy) zero-or-more operator *?, and it will match any character until it finally finds:

\\\\] ，其中字符匹配] 。注意再次逃脱。

\\], which matches the character ]. Note the escaping again.

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